From among a group of 2 men, 4 women and 6 children, 2 individuals are selected randomly. What is the probability that exactly 1 among the chosen are children?

Answer 1

prob(choosing child) = 6/12 = 1/2
prob(non-child) = 1- 1/2 = 1/2
prob(exactly one child in choosing two)
= C(2,1) (1/2)(1/2) = 1/2

or

If C stands for choosing child and N stands for non-child
we have these cases:
CC -- prob = (1/2)(1/2) = 1/4
CN -- 1/4
NC --1/4
NN -- 1/4

so prob is CN or NC = 1/4 + 1/4 = 1/2